已知数列{an}的前n项和为Sn,且Sn=1/2n^2+11/2n,数列{bn}满足b(n+2)-2b(n+1)+bn=0

题目

已知数列{an}的前n项和为Sn,且Sn=1/2n^2+11/2n,数列{bn}满足b(n+2)-2b(n+1)+bn=0
n∈N*,且b3=11,b1+b2+……+b9=153.求数列{bn}的通项公式.

答案

1.
Sn=(1/2)n^2+(11/2)n
S(n-1)=(1/2)(n-1)^2+(11/2)(n-1)
an=Sn-S(n-1)
=(1/2)n^2+(11/2)n-(1/2)(n-1)^2-(11/2)(n-1)
=(1/2)(2n-1)+11/2
=n+5
2.
b(n+2)-2b(n+1)+bn=0
则bn为等差数列
设bn=b1+(n-1)d,Tn=nb1+n(n-1)d/2
b3=b1+2d=11 ,T9=9b1+36d=153
b1=5 ,d=3
bn=5+3(n-1)=3n+2